So in this case, we can accomplish this by multiplying top and bottom
by the square root of 11:
*Den. now has a perfect square under sq. root
AND
Step 3: Simplify the fraction if needed.
*Sq. root of 121 is 11
So in this case, we can accomplish this by multiplying top and bottom
by the cube root of :
*Mult. num. and den. by cube root of
*Num. now has a perfect cube under cube root
AND
Step 3: Simplify the fraction if needed.
*Cube root of 125 y cubed is 5y
So what would the conjugate of our denominator be?
It looks like the conjugate is .
AND
Step 4: Simplify the fraction if needed.
Last revised on July 21, 2011 by Kim Seward.
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